The moment of inertia of a thin uniform rectangular plate of mass ' $m$ ', having length ' $a$ ' and width ' $b$ ' about an axis perpendicular to the plane of the plate and passing through one of its vertices is

• A

  $\frac{2}{3} \mathrm{mab}$
• B

  $\frac{1}{3} \mathrm{mab}$
• C

  $\frac{2}{3} \mathrm{~m}\left(\mathrm{a}^2+\mathrm{b}^2\right)$
• Correct answer
  D

  $\frac{1}{3} m\left(a^2+b^2\right)$

Solution

By parallel axes theorem. $\begin{aligned} & \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{md}^2 \\ & =\frac{\mathrm{m}}{12}\left(\mathrm{a}^2+\mathrm{b}^2\right)+\mathrm{m}\left(\frac{\mathrm{a}^2+\mathrm{b}^2}{4}\right) \\ & =\frac{1}{3} \mathrm{~m}\left(\mathrm{a}^2+\mathrm{b}^2\right)\end{aligned}$

Hence, the correct answer is option: 'D'.